"parameterize the plane through the point (−1,−3,1) with the normal vector ⟨−4,4,3⟩
r⃗ (s, t)=
(use s and t for the parameters in your parameterization, and enter your vector as a single vector, with angle brackets: e. g., as < 1 + s + t, s - t, 3 - t > .)"

The equation of the required plane can be obtained thus:
-4(x + 1) + 4(y + 3) + 3(z - 1) = 0
-4x - 4 + 4y + 12 + 3z - 3 = 0
4x - 4y - 3z = 5

Let x = 1, y = 2, then 4(1) - 4(2) - 3z = 5
z = (4 - 8 - 5)/3 = -9/3 = -3
Thus, point (1, 2, -3) is a point on the plane.

Let a = (a1, a2, a3) and b = (b1, b2, b3) be vectors parallel to the plane.
Then, -4a1 + 4a2 + 3a3 = 0 and -4b1 + 4b2 + 3b3 = 0
Let a1 = 2, a2 = -1, then a3 = (4(2) - 4(-1))/3 = (8 + 4)/3 = 12/3 = 4 and let b1 = -1 and b2 = 2, then b3 = (4(-1) - 4(2))/3 = (-4 - 8)/3 = -12/3 = -4
Thus a = (2, -1, 4) and b = (-1, 2, -4)

Therefore, the required parametric equation is r(s, t) = s(2, -1, 4) + t(-1, 2, -4) + (1, 2, -3) = (2s, -s, 4s) + (-t, 2t, -4t) + (1, 2, -3) = (2s - t + 1, -s + 2t + 2, 4s - 4t - 3)

That is so true, i can’t believe how true this is